The problem of the maximum value of quadratic function in the first grade of Higher Education 1. Find the maximum value of the function y = x Λ 2 + 2aX + 1, - 2 < = x < = 1. 2. Find the maximum value of the function y = x Λ 2 + 2x-1, m < = x < = m + 1. 3. Find the maximum value of the function y = - x Λ 2 + 2aX + 1-A, 0 < = x < = 1 is 2, and find a (x Λ 2 represents the square of x) (the answer should be detailed)
one
y=x^2+2ax+1=(X+A)^2+1-A^2
Because - 2 < = x < = 1
Substituting 5-4a and 2 + 2a,
When 5-4a = 2 + 2a,
A=1/2
In this case, the two formulas are equal, and the value is y = x ^ 2 + 1,
Substituting - 2 < = x < = 1,
It is known that the maximum value of the function is [0,4]
So when a is less than 1 / 2, the maximum value of the function is [2 + 2a, 5-4a]
So when it is greater than 1 / 2, the maximum value of the function is [5-4a, 2 + 2A]
2.Y=X^2+2X+1
=(X+1)^2
By substituting m < = x < = m + 1, we can see that,
(M+1)^2
The maximum value of quadratic function
Given that the square of y = 4A (x-a) (a is greater than 0), and when x is greater than or equal to a, the minimum value of S = (x-3) ^ 2 + y ^ 2 is 4, find the parameter a
This topic seems very difficult, but its expression is too complex. Let's simplify it. Substitute y ^ 2 = 4A (x-a) into S (x) = (x-3) ^ 2 + y ^ 2 to get s (x) = (x-3) ^ 2 + 4a (x-a) = x ^ 2 + (4a-6) x + (9-4a ^ 2). It's not difficult to see that this is a quadratic function. Its axis of symmetry is x = 3-2a (and its open
It is known that y = y1-y2, Y1 is proportional to the square of X, and the scale coefficient is K1
It is known that y = y1-y2, Y1 is proportional to the square of X and the scaling coefficient is K1
Y2 is proportional to x + 3, and the proportional coefficient is K2
When x = 0, y = 2
When x = 3, y = 0
Find the functional relation between Y and X. is it a linear function?
y=y1-y2
y1=k1*x^2
y2=k2*(x+3)
So y = K1 * x ^ 2-k2 * (x + 3)
Substituting two known points, K1 = - 4 / 9, K2 = - 2 / 3
So the functional relation between Y and X is y = - 4 / 9x ^ 2 + 2 / 3x + 2
Obviously it's not a linear function
Distribution of roots of quadratic function
On the equation x ^ 2 + ax + 2 = 0 of X, at least one real root is less than - 1, the range of a is obtained
It's better to do it in a variety of ways
The analysis of "at least one real root less than - 1" includes two cases: ① two real roots are less than - 1; ② one real root is less than - 1, and the other real root is not less than - 1
Solution 1 Let f (x) = x ^ 2 + ax + 2, the two real roots of the equation in the question are less than - 1, which is equivalent to
a^2-8>=0
-a/20
that is
a> = 2 times root 2 or a = 2
1-a+2>0
The solution
Double root 2
The two diagonals of diamond ABCD are 8 and 6 long respectively, point P is the moving point on diagonal AC, m and N are the midpoint of AB and BC respectively, then the minimum value of PM + PN is
5
Higher first quadratic function
It is known that the minimum value of F (x) = x & # 178; - ax + half a (a > 0) in the interval "0,1" is g (a) and the maximum value of G (a) is obtained
I'll see you tomorrow
f(x)=x²-ax+a/2=(x-a/2)²-a²/4+a/2
The function image is a parabola with the opening upward and the axis of symmetry x = A / 2
When 0
In rectangular ABCD, the diagonal lines AC and BD intersect at point O, ad = 6, CD = 8, P is the moving point on AB, PM is perpendicular to AC, PN is perpendicular to DB, and the value of PM + PN is obtained
From the meaning of the title
AO=BO=1/2BD
Because BD = under the root sign (AB & sup2; + BC & sup2;) = 10
So Ao = Bo = 5
Because the side length of △ ABO is 5,5,8
The height on the edge AB through o
On the Pythagorean theorem
The height is 3
So s △ ABO = 3 × 4 × 1 / 2 = 6
Connect op
Because s △ ABO = s △ AOP + s △ BOP=
1/2(AO×PM+BO×PN)=5/2(PM+PN)=6
So PM + PN = 2.4
If the image of quadratic function y = f (x) passes through the origin, and 1 is less than or equal to f (- 2) less than or equal to 2, 3 less than or equal to f (1) less than or equal to 4
25 / 3 less than or equal to f (2) less than or equal to 34 / 3
Because y = f (x) is a quadratic function, its equation can be set as y = f (x) = ax ^ 2 + BX + C
The image passes through the origin, i.e. the point (0,0)
So f (0) = ax0 ^ 2 + BX0 + C = 0, that is, C = 0
The quadratic function equation is reduced to y = f (x) = ax ^ 2 + BX
Another one
As shown in the figure, the diagonal of rectangle ABCD intersects at point O, P is any point of AD, PM ⊥ AC, PN ⊥ BD, if AB = 5, BC = 12, then PM + PN =?
1. Extend CD line, make parallel line of BD through a point, two lines intersect at e point, then bdea is parallelogram
2. When the NP line is extended and intersected with AE line at f point, NF is perpendicular to Ae
3. Because CD = De, ad is perpendicular to CE, so pf = PM. So PM + PN = NF
4. NF is the height of the right triangle abd, so NF = AB * BC / BD,
Calculation: ab = 5, BC = 12, so BD = 13
NF=5*12/13,
Results: NF = 60 / 13
In Cartesian coordinates, the parabola y = (- 4 / 9) x2 + (2 / 9) MX + (5 / 9) m + 4 / 3 intersects the x-axis with two points AB, point a is on the negative half axis, point B is on the positive half axis, OB = 2oa (with absolute value), and point C is the vertex of the parabola. Find the value of M
M is 4 or (- 3 / 2)